Helmholtz Free Energy

Equivalent forms

\Delta F = \Delta U - T \Delta S

dF = -S dT - P dV

Its differential dF = -SdT - PdV has S and P as natural responses — Maxwell relations follow trivially from mixed partials.

Symbol	Name	SI unit	Dimension	Range
$F$	Helmholtz energyoutput Helmholtz free energy of system	J	M·L²·T⁻²	-100000 – 100000
$U$	Internal energy Total internal energy	J	M·L²·T⁻²	0 – 1000000
$T$	Temperature Absolute temperature	K	Θ	1 – 1000
$S$	Entropy Entropy of system	J/K	M·L²·T⁻²·Θ⁻¹	0 – 1000

Unit systems

Symbol	SI	CGS	Imperial
$F$	J	erg	BTU
$U$	J	erg	BTU
$T$	K	K	°R
$S$	J/K	erg/K	BTU/°R

Where it holds

Valid for closed systems at constant T and V. For chemical reactions at constant T,P use Gibbs energy instead.

Dimensional analysis

$[F] = [U] - [T][S] = J - K\cdot (J/K) = J \checkmark$

Discovery

Hermann von Helmholtz · 1882

Helmholtz introduced this potential to clarify the distinction between 'bound' (TS) and 'free' (F) energy in chemical reactions, formalizing the work-content concept.

Try this

How much usable work can you extract from a sealed gas cylinder?

1 mol of ideal gas at 300 K expands isothermally from 1 L to 10 L. Find the change in Helmholtz energy.

Research status: stable

Real-world applications

Statistical mechanics: partition function Z gives $F = -k_B T \ln Z$
Magnetic systems and lattice models
Bomb calorimetry (constant V)
Solid-state physics — phonon free energy

Common misconceptions

F is not the same as G — F is the right potential for constant-V processes, G for constant-P
F can be negative; what matters $is \Delta F$ , not its sign
'Free' refers to extractable work, not absence of constraint

Experimental verification

Maxwell relations derived from F (e.g.,

(\partial S/\partial V)_T = (\partial P/\partial T)_V)

verified across countless equations of state. Statistical mechanics gives

F = -k_B T \ln Z

, matching all calorimetric measurements.

Derivation

Start with first law:

dU = \delta Q - PdV

.

Second law:

\delta Q \le TdS

.

So

dU \le TdS - PdV \to d(U - TS) \le -SdT - PdV

.

Define

F = U - TS

, then

dF \le -SdT - PdV

, with equality for reversible processes.

At constant T,V:

dF \le 0

— spontaneous direction is F-decreasing.

Limiting cases

T \to 0

⟶

F \to U

Internal energy dominates at low T.

Reversible isothermal process⟶

\Delta F = -W_\max

\Delta F

equals the negative of maximum extractable work.

Spontaneous at constant T,V⟶

\Delta F \le 0

F decreases for any spontaneous process at constant T,V.

What if…

What if the process is at constant pressure instead of constant volume?

Use Gibbs free energy $G = H - TS$ instead; F is the wrong potential for that constraint.

What if

T \to 0

?

$F \to U$ ; entropy term vanishes and Helmholtz energy converges to ground-state energy.

What if you know the partition function Z?

$F = -k_B T \ln Z$ — this is how statistical mechanics computes all thermodynamic properties.

1

Isothermal expansion of 1 mol ideal gas

Given ·

n:: 1
T:: 300
V 1:: 0.001
V 2:: 0.01

Find ·

\Delta F

Steps

For isothermal ideal gas, $\Delta U = 0$
$\Delta S = nR \ln (V_{2}/V_{1})$
$\Delta F = \Delta U - T\Delta S = -nRT \ln (V_{2}/V_{1}) = -1\cdot 8.314\cdot 300\cdot 2.303 \approx -5740\,\mathrm{J}$

Result ·

\Delta F = -nRT \ln (V_{2}/V_{1}) = -1\times 8.314\times 300\times \ln (10) = -5740\,\mathrm{J}

2

F of a system with U = 10 kJ, S = 100 J/K, T = 300 K

Given ·

U:: 10000
T:: 300
S:: 100

Find · F

Steps

Apply $F = U - TS$
$F = 10000 - (300)(100)$
$F = -20000\,\mathrm{J} = -20\,\mathrm{kJ}$

Result ·

F = U - TS = 10000 - 300\times 100 = -20000\,\mathrm{J}

Gibbs Free Energy→First Law of Thermodynamics→Entropy Change→